Manipal Entrance Test 2020 Math Paper

Given that,Slope of the curve $= \; \frac{y-1}{x^{2}+x}$ $\Rightarrow \; \frac{d y}{d x} = \; \frac{y-1}{x^{2}+x}$We can rewrite this as $\Rightarrow \; \frac{d y}{y-1} = \; \frac{d x}{x^{2}+x}$Integrate on both sides $\Rightarrow \int \; \frac{d y}{y-1} = \int \; \frac{d x}{x^{2}+x}$ $\Rightarrow \int \; \frac{d y}{y-1} = \int \; \frac{d x}{x(x+1)}$ $\Rightarrow \int \; \frac{dy}{y-1} = \int \left( \; \frac{1}{x} - \; \frac{1}{x+1} \right) dx$ $\Rightarrow \int \; \frac{dy}{y-1} = \int \; \frac{1}{x} dx - \int \; \frac{1}{x+1} dx$Note that this integral is in the form of $\int \; \frac{1}{a} d a = \log a$Applying this formula, we get $\Rightarrow \log (y-1) = \log x - \log (x+1) + \log C$We know that $\Rightarrow \log m + \log n = \log m n$ and $\Rightarrow \log m - \log n = \log \left( \; \frac{m}{n} \right)$Using these formulae in the above expressions we get, $\Rightarrow \log (y-1) = \log Cx - \log (x+1)$ $\Rightarrow \log (y-1) = \log \left( \; \frac{C x}{x+1} \right)$Applying Anti-log on both sides we get, $\Rightarrow y-1 = \; \frac{Cx}{x+1} \; \; \; \; \dots \text{eq}(1)$It is given that the curve passes through $(1,0)$Substituting these values of $x, y$ in the above equation we get, $\Rightarrow 0-1 = \; \frac{C \times 1}{1+1}$ $\Rightarrow -1 = \; \frac{C}{2}$ $\Rightarrow C = -2$Substituting this value of $C$ in eq(1) $\Rightarrow y-1 = \; \frac{-2 x}{x+1}$ $\Rightarrow (y-1)(x+1) = -2 x$ $\Rightarrow (y-1)(x+1) + 2x = 0$Hence, the required equation of the curve is $\Rightarrow (y-1)(x+1) + 2 x = 0$